P(C > R) = P(Z2<XY) = P([tex] - \sqrt {XY} < Z < \sqrt {XY} [/tex]) = FZ([tex]\sqrt {XY} [/tex]) - FZ(-[tex]\sqrt {XY} [/tex]), where in the second equality I used that all values of XY are positive and in the last one that FZ is absolutely continuous.

But now, since Z ~ U(0,1), its FZ(z) = z 1{0<z<1} + 1{z>1}. So FZ(-[tex]\sqrt {XY} [/tex]) = 0, because FZ(z) is 0 for all z<0.

So, I would get that P(C > R) = FZ([tex]\sqrt {XY} [/tex]) = sqrt(xy) 1{0<sqrt(xy)<1} + 1{sqrt(xy)>1}. XY is bounded by 1, so sqrt(xy) can't be >1. Then this is sqrt(xy) 1{0<sqrt(xy)<1}.